Q. A and B together can do a piece of work in 10 days, while B and C together can do it in 12 days and C and A together in 15 days. In how many days C would complete it alone?
Answer: 40 days
Notes: (A + B)'s 1 day's work $latex = \frac{1}{10}&s=1$......(I)
(B + C)'s 1 day's work $latex = \frac{1}{12}&s=1$......(II)
(C + A)'s 1 day's work $latex = \frac{1}{15}&s=1$......(III)
On adding all the equations,
2 (A +B + C)'s 1 day's work $latex = \frac{1}{10} + \frac{1}{12} + \frac{1}{15} = \frac{6+5+4}{60} = \frac{1}{4}&s=1$
∴ (A + B + C)'s 1 day's work $latex = \frac{1}{8}&s=1$......(IV)
Now, C's 1 day's work = (A + B + C)'s 1 day's work - (A + B)'s 1 day's work
$latex = \frac{1}{8}-\frac{1}{10}=\frac{5-4}{40}=\frac{1}{40}&s=1$
Hence C will finish the work in 40 days. So option [C] is correct answer.
